MHB Limit of f(x+y)=f(x)+f(y) near 0

[alyafey22]

Let $f: \mathbb{R}\to \mathbb{R}$ be such that $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$ . Assume the limit of $\lim_{x\to 0}f(x)=L$ then prove that $L=0$.

Here is my attempt:

We first choose an arbitrary $$\epsilon >0$$ then we choose $\delta$ such that $$\text{Max}(|x|,|y|,|x+y|)<\delta $$ so we can say that $$|f(y)-L|<\frac{\epsilon}{2} $$ and $$|f(x+y)-L|<\frac{\epsilon}{2} $$

Since we have the following

$$|f(x+y)-L|=|f(x)+f(y)-L| \geq |f(x)|-|f(y)-L|$$

$$|f(x)|\leq |f(x+y)-L|+|f(y)-L|<\frac{\epsilon}{2} +\frac{\epsilon}{2} =\epsilon $$ for the chosen $$\delta $$ by existence of limit of $$f$$.

So we deduce that

$$0<|x|<\delta $$ implies $$|f(x)|<\epsilon $$

since $$\epsilon >0$$ is arbitrary we have $$\lim_{x \to 0}f(x)=0 \,\,\, \square $$.

So what you think guys ?

 

[zzephod]

Let $f: \mathbb{R}\to \mathbb{R}$ be such that $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$ . Assume the limit of $\lim_{x\to 0}f(x)=L$ then prove that $L=0$.

Here is my attempt:

We first choose an arbitrary $$\epsilon >0$$ then we choose $\delta$ such that $$\text{Max}(|x|,|y|,|x+y|)<\delta $$ so we can say that $$|f(y)-L|<\frac{\epsilon}{2} $$ and $$|f(x+y)-L|<\frac{\epsilon}{2} $$

Since we have the following

$$|f(x+y)-L|=|f(x)+f(y)-L| \geq |f(x)|-|f(y)-L|$$

$$|f(x)|\leq |f(x+y)-L|+|f(y)-L|<\frac{\epsilon}{2} +\frac{\epsilon}{2} =\epsilon $$ for the chosen $$\delta $$ by existence of limit of $$f$$.

So we deduce that

$$0<|x|<\delta $$ implies $$|f(x)|<\epsilon $$

since $$\epsilon >0$$ is arbitrary we have $$\lim_{x \to 0}f(x)=0 \,\,\, \square $$.

So what you think guys ?

Observe that: \(f(x)=2f(x/2)\), then as \(\lim_{x \to 0} f(x) = \lim_{x \to 0} f(x/2)\) we have \(L=2L\) etc

.

 

[Opalg]

So what you think guys ?

Your proof looks fine (though it's not as slick as zzephod's).

 

[Saitama]

Not a rigorous proof but f(x)=kx (k is a constant) satisfies the given functional relation so the limit is zero.

 

[alyafey22]

Not a rigorous proof but f(x)=kx (k is a constant) satisfies the given functional relation so the limit is zero.

I don't get your point .

 

[Saitama]

I don't get your point .

I am sorry if I used the wrong words. I have no knowledge of those epsilon and delta proofs you deal with. The functional relation "f(x+y)=f(x)+f(y)" is a common one for me (I mean I see it frequently while doing problems). The function f(x)=kx satisfies the above relation i.e f(x+y)=kx+ky and f(x)+f(y)=kx+ky.
So $\lim_{x \rightarrow 0} kx=0$. You may ignore my post if you wish to. :)

 

[Nono713]

I am sorry if I used the wrong words. I have no knowledge of those epsilon and delta proofs you deal with. The functional relation "f(x+y)=f(x)+f(y)" is a common one for me (I mean I see it frequently while doing problems). The function f(x)=kx satisfies the above relation i.e f(x+y)=kx+ky and f(x)+f(y)=kx+ky.
So $\lim_{x \rightarrow 0} kx=0$. You may ignore my post if you wish to. :)

I think what Zaid means is that while you showed that this particular function works, that does not really show that all functions that satisfy $f(x + y) = f(x) + f(y)$ meet the given condition (though in this case they happen to). As an example: just because you are an internet user, and also an MHB member, doesn't mean all internet users are also MHB members.​

 

[Saitama]

I think what Zaid means is that while you showed that this particular function works, that does not really show that all functions that satisfy $f(x + y) = f(x) + f(y)$ meet the given condition (though in this case they happen to). As an example: just because you are a internet user, and also an MHB member, doesn't mean all internet users are also MHB members.​

Thanks Bacterius!

I am interested to know if there are any other functions which satisfy the given functional relation. Do you have any ideas? :)

 

[Nono713]

Thanks Bacterius!

I am interested to know if there are any other functions which satisfy the given functional relation. Do you have any ideas? :)

Hmm. There are discontinuous functions which satisfy the relation (see Cauchy's functional equation - Wikipedia, the free encyclopedia) but Zaid's claim requires continuity at $x = 0$ for the limit to exist. So they may or may not apply to the problem, I am not sure. In any case, it is still important to verify that the claim still holds over (or applies to) all such functions, otherwise the proof does not quite follow.

 

[I like Serena]

Thanks Bacterius!

I am interested to know if there are any other functions which satisfy the given functional relation. Do you have any ideas? :)

How about:
$$f(x) = \begin{cases}
3x & \text{ if } x \in \mathbb Q \\
-2x & \text{ otherwise }
\end{cases}$$

 

[Saitama]

How about:
$$f(x) = \begin{cases}
3x & \text{ if } x \in \mathbb Q \\
-2x & \text{ otherwise }
\end{cases}$$

Sorry for the dumb question, how do I evaluate f(x+y), f(x) and f(y) here?

 

[alyafey22]

Define the Laplace transform as follows

$$F(s)=\mathcal{L}(f(t))=\int^{\infty}_0 e^{-st}\, f(t)\, dt $$

Take the space of all functions that have a Laplace transform then

$$\mathcal{L}(x(t)+y(t))=\int^{\infty}_0 e^{-st}\, (x(t)+y(t))\, dt = \int^{\infty}_0 e^{-st}\, x(t)\, dt \, +\,\int^{\infty}_0 e^{-st}\, y(t)\, dt \,= \mathcal{L}(x(t))+\mathcal{L}(y(t)) $$

 

[I like Serena]

How about:
$$f(x) = \begin{cases}
3x & \text{ if } x \in \mathbb Q \\
-2x & \text{ otherwise }
\end{cases}$$

Sorry for the dumb question, how do I evaluate f(x+y), f(x) and f(y) here?

Sorry, I have just realized that my suggestion does not work.
$$f(1+\pi) = -2(1+\pi) = -2 -2\pi$$
$$f(1) + f(\pi) = 3 \cdot 1 + (-2)\cdot \pi = 3 - 2\pi$$
Obviously they are not the same.

EDIT: The following function does work though.
$$f(x) = \begin{cases}
0 & \text{ if } x \in \mathbb Q \\
2x & \text{ otherwise }
\end{cases}$$

 

[I like Serena]

Define the Laplace transform as follows

$$f(s)=\mathcal{L}(f(t))=\int^{\infty}_0 e^{-st}\, f(t)\, dt $$

Take the space of all functions that have a Laplace transform then

$$\mathcal{L}(x(t)+y(t))=\int^{\infty}_0 e^{-st}\, (x(t)+y(t))\, dt = \int^{\infty}_0 e^{-st}\, x(t)\, dt \, +\,\int^{\infty}_0 e^{-st}\, y(t)\, dt \,= \mathcal{L}(x(t))+\mathcal{L}(y(t)) $$

Hmm, but $\mathcal{L}$ isn't a function of $\mathbb R \to \mathbb R$ is it?

 

[alyafey22]

We can have a generalization for all linear operators

$$L(f+g)=L(f)+L(g) $$

$$L(k\, f) = k\, L(f) \,\,\,\, $$ where k is constant

 

[Fernando Revilla]

Let $f: \mathbb{R}\to \mathbb{R}$ be such that $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$ . Assume the limit of $\lim_{x\to 0}f(x)=L$ then prove that $L=0$.

Another way: $L=\displaystyle\lim_{x\to 0}f(x)$ implies $\displaystyle\lim_{n\to +\infty}f\left(\frac{1}{n}\right)=L.$

But (easily proved), $\displaystyle f\left(\frac{1}{n}\right)=\frac{f(1)}{n},$ so $L=\displaystyle\lim_{n\to +\infty}\frac{f(1)}{n}=0.$

 

[alyafey22]

Hmm, but $\mathcal{L}$ isn't a function of $\mathbb R \to \mathbb R$ is it?

Take the space of all constant functions in $$\mathbb{R}$$ then $$x(t) =a , y(t) = b $$

$$\mathcal{L}(a+b)=\mathcal{L}(a)+\mathcal{L}(b)$$

 

[I like Serena]

Take the space of all constant functions in $$\mathbb{R}$$ then $$x(t) =a , y(t) = b $$

$$\mathcal{L}(a+b)=\mathcal{L}(a)+\mathcal{L}(b)$$

Ah, but when you write $\mathcal{L}(a)$, with $a$ you mean a constant function, which is an element of $\mathbb R \to \mathbb R$ given by $t \mapsto a$, which is still not an element of $\mathbb R$.

More importantly, $\mathcal{L}(a)$ is also a function given by $$s \mapsto \frac a s$$, which is not an element of $\mathbb R$.

 

[Opalg]

EDIT: The following function does work though.
$$f(x) = \begin{cases}
0 & \text{ if } x \in \mathbb Q \\
2x & \text{ otherwise }
\end{cases}$$

Sorry, that doesn't work either. For example, if $x=\pi$ and $y=4-\pi$ then $f(x)+f(y) = 2\pi + 2(4-\pi) = 8$ but $f(x+y) = f(4) = 0.$

To construct a discontinuous function on $\mathbb{R}$ that satisfies $f(x+y) = f(x) + f(y)$, you need to define it on the elements of a Hamel basis (see the link in Bacterius's comment http://mathhelpboards.com/analysis-50/limit-f-x-y-%3Df-x-f-y-near-0-a-7582.html#post34576). The existence of a Hamel basis depends in turn on the axiom of choice. I am fairly sure that it is not possible to find an example of a discontinuous $f$ without using the axiom of choice.